题面

Vjudge

Vjudge

Sol

求一个串不同子串的个数

每个子串一定是某个后缀的前缀，也就是求所有后缀不同前缀的个数

每来一个后缀\(suf(i)\)就会有，\(len-sa[i]+1\)的新的前缀，又由于有\(height\)个重复的，那么就是\(len-sa[i]+1-height\)的贡献

两个就只有数组大小的区别

# include <bits/stdc++.h>

# define IL inline

# define RG register

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(1010);

int n, a[_], sa[_], rk[_], tmp[_], height[_], t[_], mx, T;

char s[_];

IL bool Cmp(RG int x, RG int y, RG int k){  return tmp[x] == tmp[y] && tmp[x + k] == tmp[y + k];  }

IL void Suffix_Sort(){

	RG int m = mx;

	for(RG int i = 0; i <= m; ++i) t[i] = 0;

	for(RG int i = 1; i <= n; ++i) ++t[rk[i] = a[i]];

	for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];

	for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;

	for(RG int k = 1; k <= n; k <<= 1){

		RG int l = 0;

		for(RG int i = n - k + 1; i <= n; ++i) tmp[++l] = i;

		for(RG int i = 1; i <= n; ++i) if(sa[i] > k) tmp[++l] = sa[i] - k;

		for(RG int i = 0; i <= m; ++i) t[i] = 0;

		for(RG int i = 1; i <= n; ++i) ++t[rk[tmp[i]]];

		for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];

		for(RG int i = n; i; --i) sa[t[rk[tmp[i]]]--] = tmp[i];

		swap(rk, tmp); rk[sa[1]] = l = 1;

		for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;

		if(l >= n) break;

		m = l;

	}

	for(RG int i = 1, h = 0; i <= n; ++i){

		if(h) --h;

		while(a[i + h] == a[sa[rk[i] - 1] + h]) ++h;

		height[rk[i]] = h;

	}

}

IL int Calc(){

	RG int ans = n - sa[1] + 1;

	for(RG int i = 2; i <= n; ++i) ans += n - sa[i] + 1 - height[i];

	return ans;

}

int main(RG int argc, RG char* argv[]){

	scanf("%d", &T);

	while(T--){

		scanf(" %s", s + 1);

		n = strlen(s + 1); mx = 0;

		for(RG int i = 1; i <= n; ++i) a[i] = s[i], mx = max(mx, a[i]);

		Suffix_Sort();

		printf("%d\n", Calc());

	}

    return 0;

}